视频BV1Fy4y1S7uj题12.解析
f（x）关于直线x＝π/2对称
且在[-π/2，π/2]单增
故（π-a-π/2）²＜（π/2+a-π/2）²
且π-a∈[-π/2，3π/2]
且π/2+a∈[-π/2，3π/2]
即a∈（π/4，π]
ps.
本题考察函数单调性与对称性
f（x）关于直线x＝π/2对称
且在[-π/2，π/2]单增
故（π-a-π/2）²＜（π/2+a-π/2）²
且π-a∈[-π/2，3π/2]
且π/2+a∈[-π/2，3π/2]
即a∈（π/4，π]
ps.
本题考察函数单调性与对称性
